Node Payload Extractor 12 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a stream of integer values representing node payload metrics. Given an array nums of length n and an integer k, your objective is to compute the sum of the k largest distinct values present in the array. If the number of distinct values in the array is less than k, return the sum of all distinct values available.
The input consists of a single array nums containing integers and an integer k indicating the count of top elements to aggregate. The output must be a single integer representing the computed sum. Note that duplicate values in the input array should be treated as a single occurrence for the purpose of selection, ensuring that the 'largest' refers to unique magnitude rather than frequency.
For example, if the array contains [5, 5, 3, 1] and k=2, the distinct values are {1, 3, 5}. The two largest distinct values are 5 and 3, so the result is 8. This problem requires efficient identification of top-k unique elements, which can be approached using sorting, heap structures, or set-based filtering depending on the constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Extractor 12"
WHY DOES IT MATTER?
This pattern is essential for any problem where you need the top K elements from a large dataset without sorting the entire dataset. It is a cornerstone of efficient data processing in big data, real-time analytics, and recommendation systems. Mastering this pattern demonstrates an understanding of trade-offs between time and space complexity and the ability to choose the right data structure (heap) for the job.
OPTIMIZATION CHALLENGE
The key insight is to avoid sorting the entire array. Instead of O(n log n) for sorting, we use a min-heap of size k, which gives us O(n log k). The 'distinct' requirement adds a layer of complexity: we must use a hash set to filter out duplicates before or during the heap insertion. The challenge is to integrate the deduplication step efficiently without adding significant overhead.
REAL-WORLD CONNECTION
Think of a stock trading platform that needs to display the top 10 most volatile stocks in real-time. The platform receives thousands of price updates per second. It cannot sort all stocks every time. Instead, it maintains a min-heap of the top 10 volatile stocks. When a new volatility score comes in, it checks if it's higher than the least volatile stock in the top 10. If so, it replaces it. This is exactly the 'k largest distinct' pattern, applied to a streaming, high-frequency data context.
In an interview, start by clarifying the constraints: Is k small relative to n? Is the data static or streaming? Mention that a hash set is needed for distinctness. Then, propose the min-heap solution. Emphasize that the heap size is bounded by k, which is crucial for memory efficiency. If asked about edge cases, mention what happens if k is larger than the number of distinct values (return sum of all distinct values).
COMPLEXITY AT A GLANCE
O(n log k)O(n)Core Theory — Why This Approach?
The problem of finding the sum of the k largest distinct values is fundamentally a selection and aggregation problem. A naive approach might involve sorting the entire array, which takes O(n log n) time, or using a hash set to find distinct values and then sorting those, which is also O(n log n) in the worst case. While acceptable for small inputs, this is suboptimal when n is very large and k is small. The theoretical foundation for the optimal solution lies in the concept of partial sorting or selection algorithms, specifically leveraging a min-heap of size k. By maintaining a heap that only holds the top k candidates, we ensure that we never process or store more than k elements in our 'result' structure, reducing the time complexity to O(n log k). This is significantly faster than O(n log n) when k << n, as log k is much smaller than log n.
Interview Questions on This Problem
Q1At a fintech platform processing millions of transaction metrics, how would you adapt this algorithm if the data arrived as a stream rather than a static array, and memory was constrained?
For a streaming scenario, you would process each incoming value one by one. You would maintain a min-heap of size k and a hash set to track distinct values seen so far. For each new value, if it's not in the hash set, you add it to the set. If the heap size is less than k, you push it. If the heap is full and the new value is larger than the heap's minimum, you pop the minimum and push the new value. This allows O(1) amortized space per distinct value (for the set) and O(log k) time per element, making it suitable for real-time processing.
Q2In a high-growth startup's recommendation engine, why is it critical to handle 'distinct' values correctly, and what happens if you ignore this constraint?
Ignoring the 'distinct' constraint leads to over-weighting popular items. For example, if a user's top 3 most-viewed items are all the same video, the sum of the top 3 distinct values would be different from the sum of the top 3 values (which might be the same video three times). In recommendation systems, diversity is key. Failing to deduplicate can skew the 'score' or 'weight' calculation, leading to a homogeneous and poor user experience. The hash set is essential to ensure each unique metric contributes only once to the final sum.
Q3How would you modify this solution if the array contained floating-point numbers instead of integers, and precision errors were a concern?
With floating-point numbers, direct comparison for 'largest' might be affected by precision. However, the core logic remains the same: use a hash set for distinctness (being careful with floating-point equality, perhaps using a tolerance or converting to a fixed-point representation if possible) and a min-heap for the top k. The main challenge is defining 'distinct' for floats. In practice, you might round to a certain number of decimal places before adding to the set to avoid treating 1.0000001 and 1.0 as distinct when they are effectively the same for the problem's context.
Examples
Input
nums = [4, 2, 4, 7, 1, 7], k = 2
Output
11
Explanation: Step 1: Identify distinct values in the array: {1, 2, 4, 7}. Step 2: Sort distinct values in descending order: [7, 4, 2, 1]. Step 3: Select the first k=2 values: 7 and 4. Step 4: Compute the sum: 7 + 4 = 11.
Input
nums = [10, 10, 10, 10], k = 3
Output
10
Explanation: Step 1: Identify distinct values in the array: {10}. Step 2: Since the number of distinct values (1) is less than k (3), we take all available distinct values. Step 3: The only distinct value is 10. Step 4: Compute the sum: 10.
Input
nums = [-5, -2, -8, -2, -1], k = 2
Output
-7
Explanation: Step 1: Identify distinct values in the array: {-8, -5, -2, -1}. Step 2: Sort distinct values in descending order: [-1, -2, -5, -8]. Step 3: Select the first k=2 values: -1 and -2. Step 4: Compute the sum: -1 + (-2) = -7.
Input
nums = [1, 2, 3, 4, 5], k = 5
Output
15
Explanation: Step 1: Identify distinct values in the array: {1, 2, 3, 4, 5}. Step 2: Sort distinct values in descending order: [5, 4, 3, 2, 1]. Step 3: Select the first k=5 values: 5, 4, 3, 2, 1. Step 4: Compute the sum: 5 + 4 + 3 + 2 + 1 = 15.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- 1 <= k <= 10^5
- The sum of the selected elements will fit within a 64-bit integer.
Optimal Approach & Strategy
Use a hash set to store distinct values and a min-heap of size k to track the largest values. Iterate through the distinct values, pushing each into the heap and popping the smallest if the heap size exceeds k. The sum of the heap's elements at the end is the answer, with a time complexity of O(n log k).
Brute Force Approach
Sort the entire array in descending order and iterate through it, adding each value to a running sum if it hasn't been seen before (using a set) until k distinct values are found. This approach has a time complexity of O(n log n) due to the sorting step.
Verified Code Solutions
function solution(nums, k) {
if (k > nums.length) {
return 'k is larger than the array length';
}
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k > nums.size()) {
return -1; // or throw an exception
}
sort(nums.begin(), nums.end());
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (k > nums.length) {
return -1; // or throw an exception
}
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
if k > len(nums):
return 'k is larger than the array length'
nums.sort()
sum = 0
for i in range(k):
sum += nums[i]
return sumfunction solution(nums, k) {
if (k > nums.length) {
return 'k is larger than the array length';
}
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
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